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Let $C_1, C_2$ be two circles touching each other externally at the point $A$ and let $A B$ be the diameter of circle $C_1$. Draw a secant $B A_3$ to circle $C_2$, intersecting circle $C_1$ at a point $A_1(\neq A)$, and circle $C_2$ at points $A_2$ and $A_3$. If $B A_1=2, B A_2=3$ and $B A_3=4$, then the radii of circles $C_1$ and $C_2$ are respectively
$\frac{\sqrt{30}}{5}, \frac{3 \sqrt{30}}{10}$
$\frac{\sqrt{5}}{2}, \frac{7 \sqrt{5}}{10}$
$\frac{\sqrt{6}}{2}, \frac{\sqrt{6}}{2}$
$\frac{\sqrt{10}}{3}, \frac{17 \sqrt{10}}{30}$
Solution
(a)
Given,
$A B$ is diameter of circle $C_1$.
$B A_1=2$
$B A_2=3$
$B A_3=4$
Let radius of circle $C_1=r_1$ and radius of circle $C_2=r_2$
$\therefore B A=2 r_1 \text { and } A C=2 r_2$
$\Rightarrow B M=\frac{1}{2} B A_1=1$
$\Rightarrow B N=B A_2+\frac{1}{2} A_2 A_3$
$=3+\frac{1}{2}=\frac{7}{2}$
In $\triangle B M P$ and $\triangle B N Q$
$\frac{B M}{B N} =\frac{B P}{B Q}$
$\frac{1}{7 / 2} =\frac{r_1}{2 r_1+r_2}$
$2 r_2 =3 r_1$
Now, $\quad B A_2 \times B A_3=B A \times B C$
$3 \times 4=2 r_1\left(2 r_1+2 r_2\right)$
$12=4\left(r_1^2+r_1 r_2\right)$
$r_1^2+r_1 r_2=3$
From Eqs.$(i)$ and $(ii)$, we get
$r_1=\sqrt{\frac{6}{5}}=\frac{\sqrt{30}}{5}$ and $r_2=\frac{3 \sqrt{30}}{10}$
